Question

我在buildForm中有以下代码，我想获得Collection<Object>的翻译，但我找不到方法。有可能吗？

if(Groups.GroupBy(e=>e.GroupName).Count()>1)
{
 //...
}

Answer 1

尝试明确设置translation_domain：

$builder->add('subject', TextType::class, array(
    // ...
    'translation_domain' => 'messages',
))

您还可以通过将此方法添加到：{/ p>来为整个FormType执行此操作

public function setDefaultOptions(OptionsResolverInterface $resolver)
{    
    $resolver->setDefaults(array(
        'translation_domain' => 'messages'
    ));
}

如果它不起作用，请通过注册as a service translator注入FormType {/ 1}}：

# services.yml
services:
    app.form.type.abuse:
        class: AppBundle\Form\Type\AbuseType
        arguments: [ "@translator" ]
        tags:
            - { name: form.type }

直接致电翻译：

use Symfony\Component\Translation\TranslatorInterface;

class AbuseType extends AbstractType
{
    public function __construct(TranslatorInterface $translator)
    {
        $this->translator = $translator;
    }

    // ...

    $builder->add('subject', TextType::class, array(
        'label' => 'label.subject',
        'data' => $this->translator->trans('report_abuse', array(), 'messages'),
        'disabled' => 'true',
    ))
}

Answer 2

所以这就是我所做的。我已经在滥用类中注入了主题，因为默认情况下它的值将用于构建表单。我还将字段设置为readonly而不是disabled，这会导致字段未提交。

//FormController.php
    class FormController extends Controller
    {
        //...
        public function abuseAction(Request $request)
        {
            $subject = $this->get('translator')->trans('report_abuse');
            $abuse = new Abuse($subject);
           //...
        }
        //...
    }

//Abuse.php
    class Abuse
    {
        //...

        public function __construct($subject)
        {
            $this->setSubject($subject);
        }

    //...
    }

//AbuseType.php
class AbuseType extends AbstractType
{
    //...

    $builder->add('subject', TextType::class, array(
        'label' => 'label.subject',
        'attr' => array(
            'readonly' => true,
        )))

    //...
}

Symfony如何在表单中转换数据值

2 个答案: